Is the ring of global functions on an integral scheme an integral domain (if the scheme is affine then it is try by definition so we are interested in non-affine schemes)? It is necessarily a reduced ring (https://stacks.math.columbia.edu/tag/01OL) so the question is whether it is irreducible.
Is the ring of global functions on an integral scheme integral
algebraic-geometry
Best Answer
If $X$ is any reduced scheme and $U$ is a dense open subset, then the restriction map $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ is injective. Indeed, an element of the kernel vanishes in the fiber at every point of $U$, but the set where a section of $\mathcal{O}_X$ vanishes in the fiber is closed, so it must vanish in the fiber everywhere. Since $X$ is reduced, a function which vanishes in the fiber everywhere must be $0$.
In particular, if $X$ is integral, we can consider $\mathcal{O}_X(X)$ as a subring of the domain $\mathcal{O}_X(U)$ for any nonempty affine open $U$, and so $\mathcal{O}_X(X)$ is also a domain.